impl Solution {
    fn first_missing_positive(mut nums: Vec<i32>) -> i32 {
        // The length of the input array.
        let n = nums.len() as i32;

        // Iterate over each element in the array.
        for i in 0..n {
            // The while loop is used to place the current element at its correct position.
            // We keep swapping until:
            // 1. The current element is out of the range [1, n].
            // 2. The current element is already in its correct position.
            //    This is checked by comparing the current element with the element at its 'ideal' index.
            while nums[i as usize] > 0
                && nums[i as usize] <= n
                && nums[nums[i as usize] as usize - 1] != nums[i as usize]
            {
                // Swap the current element with the element at its 'ideal' position.
                let num = nums[i as usize]; // Store nums[i as usize] in a temporary variable
                nums.swap(i as usize, num as usize - 1);
            }
        }

        // After rearranging, iterate through the array again.
        for i in 0..n {
            // If the value at index i is not i + 1, it means i + 1 is the missing positive integer.
            // This is because in the ideal arrangement, each integer x should be at index x - 1.
            if nums[i as usize] != i + 1 {
                return i + 1;
            }
        }

        // If all numbers are in their correct positions and no positive integer is missing in the range [1, n],
        // then the smallest missing positive integer is n + 1.
        n + 1
    }
}